What is the answer to 2^{3x-7}=5^{x-4} ?

The answer to 2^{3x-7}=5^{x-4} is x=(-4ln(5)+7ln(2))/(3ln(2)-ln(5))

Solve for x: 2^{3x-7}=5^{x-4}

The solution is x=(-4ln(5)+7ln(2))/(3ln(2)-ln(5))

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2^{3 x - 7} = 5^{x - 4}

x = \frac{- 4 ln ( 5 ) + 7 ln ( 2 )}{3 ln ( 2 ) - ln ( 5 )}

Decimal Notation

x = - 3.37384 \dots

Solution steps

2^{3 x - 7} = 5^{x - 4}

Apply exponent rules

(3 x - 7) ln (2) = (x - 4) ln (5)

Solve

(3 x - 7) ln (2) = (x - 4) ln (5) : x = \frac{- 4 ln ( 5 ) + 7 ln ( 2 )}{3 ln ( 2 ) - ln ( 5 )}

x = \frac{- 4 ln ( 5 ) + 7 ln ( 2 )}{3 ln ( 2 ) - ln ( 5 )}

6 4^{x - 4} = 1 6^{x - 3}

3^{2 x} - 3^{x - 1} = 3^{x + 1} - 1

y, x^{y} = 0

3^{2 x - 7} = 27

2^{x^{3}} = 3 2^{x}

What is the answer to 2^{3x-7}=5^{x-4} ?
The answer to 2^{3x-7}=5^{x-4} is x=(-4ln(5)+7ln(2))/(3ln(2)-ln(5))
Solve for x: 2^{3x-7}=5^{x-4}
The solution is x=(-4ln(5)+7ln(2))/(3ln(2)-ln(5))